Prime Factorization and Domination in the Hierarchical Product of Graphs

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1 Prime Factorization and Domination in the Hierarchical Product of Graphs S. E. Anderson 1, Y. Guo 2, A. Rubin 2 and K. Wash 2 1 Department of Mathematics, University of St. Thomas, St. Paul, MN Department of Mathematics, Trinity College, Hartford, CT September 23, 2015 Abstract. In 2009, Barrière et al. introduced the generalized hierarchical product of graphs. This is a generalization of the Cartesian product of graphs in that we can represent a Cartesian product as a generalized hierarchical product. It is known that every connected graph has a unique prime factor decomposition with respect to the Cartesian product. We generalize this result to show that connected graphs indeed have a unique prime factor decomposition with respect to the generalized hierarchical product. We also give preliminary results on the domination number of generalized hierarchical products. Keywords: Keywords: hierarchical product, Cartesian product, prime factor decomposition AMS classification: 05C70, 05C76 1. Introduction The most appealing aspect of the generalized hierarchical product of graphs is that it is a generalization of the well-studied Cartesian product of graphs. In particular, within a Cartesian product graph there exist an interesting set of subgraphs which possess many of the algebraic properties that the Cartesian product enjoys. In the literature for product graphs, Godsil and McKay [5] first defined a rooted product graph in 1978; a class of graphs which is a proper subset of the class of all generalized hierarchical product graphs. Rooted product graphs arise throughout the literature in various contexts, some of which can be seen in [1, 4, 8, 9]. Then in 2009, Barrière, Dalfó, Fiol, and Mitjana [2] introduced the generalized hierarchical product in an effort to model scale-free networks such as the World Wide Web, some metabolic networks, and telephone graphs [13, 14]. The generalized hierarchical product is actually a generalization of the rooted product graph as well as a generalization of the Cartesian product. 1

2 2 Sabidussi [11] and Vizing [15] showed independently that not only does every graph have a prime factor decomposition with respect to the Cartesian product, but that this prime factorization is unique for connected graphs. An alternate proof of this property can also be found in [6]. As with any meaningful attempt at generalizing a set that exhibits some algebraic structure, we must investigate whether each of the important algebraic properties of the Cartesian product carries over to the generalized hierarchical product, heretofore referred to as simply the hierarchical product. Barrière et al. [2, 12] showed that indeed the hierarchical product is associative, but it is not commutative. We now ask whether this new product retains the very important property from the Cartesian product: unique prime factorization. For if this product does not exhibit unique prime factorization, then the study of a large class of parameters within such a product would become meaningless as we prefer to relate the value of any invariant for a product graph to the values for the invariant in the product s underlying factor graphs. The remainder of this paper is organized as follows. Section 1.1 is dedicated to notation and preliminary results. In Section 2, we show that although prime factor decomposition with respect to the hierarchical product need not be unique in the case of disconnected graphs, it is in fact unique in the class of connected graphs. In Section 3, we give preliminary results on the domination number of specific hierarchical products Preliminaries. We consider only finite, simple, and undirected graphs. Given a graph G, we let V (G) represent the vertex set of G and E(G) represent the edge set of G. The Cartesian product of graphs G and H, denoted GH, is the graph whose vertex set is V (G) V (H), whereby two vertices (u 1, u 2 ) and (v 1, v 2 ) are adjacent if u 1 v 1 E(G) and u 2 = v 2, or u 1 = v 1 and u 2 v 2 E(H). Figure 1 depicts the Cartesian product of K 2 and K 3. Given graphs G 1,..., G n and nonempty vertex subsets U i V (G i ) for 1 i n 1, the generalized hierarchical product, denoted G = G 1 (U 1 ) G n 1 (U n 1 ) G n, is the graph with vertex set V (G 1 ) V (G n ) and adjacencies (y 1, x 2,..., x n ) if x 1 y 1 E(G 1 ) (x 1, y 2, x 3,..., x n ) if x 2 y 2 E(G 2 ) and x 1 U 1 (x 1,..., x n 1, x n ).. (x 1,..., x n 1, y n ) if x n y n E(G n ) and x i U i, 1 i n 1. Figure 2 depicts the hierarchical product K 3 (U 1 ) K 3 (U 2 ) K 2, where V (K n ) = {0,..., n 1} for n {2, 3} and U 1 = U 2 = {0, 2}. The role of the Cartesian product is quite important in the proof technique used in later sections. For this reason, we make a natural association between a given hierarchical product and a Cartesian product.

3 3 Figure 1. K 2 K Figure 2. K 3 (U 1 ) K 3 (U 2 ) K 2 where U 1 = U 2 = {0, 2} Definition 1. Given a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n, the Cartesian product associated with G is the graph G = G 1 G n. In general, given graphs G 1,..., G n and graph product operation, for any index 1 i n there is a projection map p i : G 1 G n 1 G n G i defined as p i ((x 1,..., x n )) = x i. We call x i the i th coordinate of the vertex (x 1,..., x n ). When represents the hierarchical product operation, then we must also choose subsets U i V (G i ) for 1 i n 1 so that p i : G 1 (U 1 ) G n 1 (U n 1 ) G n G i. We refer to G i as a factor with respect to its graph product G 1 G n. By definition, when G = G 1 G n, if (x 1,..., x n )(y 1,..., y n ) is an edge of E(G), then x i = y i or x i y i E(G i ) for each 1 i n. For this reason, each projection p i is a weak homomorphism. Fix j {1,..., n} and let a = (a 1,..., a n ) V (G). The G j -layer through a is the induced subgraph G a j = {x V (G) p i (x) = a i for i j}.

4 4 Notice that if a i U i for 1 i < j, then G a j is connected. Moreover, the restriction p i : G a j G j is an isomorphism so G a j = G j. On the other hand, if a i U i for some i {1,..., j 1}, then G a j is a totally disconnected graph. As mentioned above, in [2], Barrière et al. show that the hierarchical product is associative. Thus, it suffices to study the case of two factors when appropriate. A graph is prime with respect to a given graph product if it is nontrivial and cannot be represented as the product of two nontrivial graphs. Letting K 1 represent the graph consisting of a single vertex, then a nontrivial graph G is prime with respect to the hierarchical product if G = G 1 (U 1 ) G 2 implies that G 1 or G 2 is K 1. The result that every graph has a prime factor decomposition with respect to the Cartesian product was shown in [6, p. 65]. This same argument also shows that every graph has a prime factor decomposition with respect to the hierarchical product. Proposition 2 ([6], p. 65). Every nontrivial graph G has a prime factor decomposition with respect to the hierarchical product. The number of prime factors is at most log 2 V (G). 2. Prime Factor Decomposition We first consider prime factorizations with respect to the hierarchical product when at least one of the factor graphs is disconnected. Prime factorization is not unique with respect to the Cartesian product of two graphs when at least one of the factor graphs is disconnected. Consequently, it is not unique with respect to the hierarchical product as the Cartesian product is a specific type of hierarchical product. We choose a graph which cannot be represented as a Cartesian product to exemplify this fact. Theorem 3. Prime factorization is not unique for the hierarchical product in the class of possibly disconnected simple graphs. Proof. First, note that (K 1 + K 2 )(U 1 ) K 2 = K2 (W 1 ) (K 1 + K 1 + K 1 ), where U 1 is the isolated vertex of K 1 + K 2 and W 1 =. Figure 3 illustrates the two factorizations given, where V (K 1 + K 2 ) = V (K 1 + K 1 + K 1 ) = {0, 1, 2} and V (K 2 ) = {0, 1}. Notice that the cardinality of each of the factors K 2, K 1 + K 2, and K 1 +K 1 +K 1 is prime. It follows that each of the factors in both prime factorizations is prime. Next we show the uniqueness of prime factorization with respect to the hierarchical product for connected graphs. Throughout the remainder of this section, we assume that all graphs are connected. Note that a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n is connected if and only if each G i is connected for all 1 i n and each U j for all 1 j n 1. We generalize the method of proof given in [6, p ] which was used to prove the uniqueness of prime

5 (a) (K 1 + K 2 )(U 1 ) K 2 where U 1 = {0} (b) K 2 ( ) (K 1 + K 1 + K 1 ) Figure 3. Two different prime factorizations of 3K 2 factorization with respect to the Cartesian product. Furthermore, we shall focus on connected subgraphs of a given hierarchical product. Of particular importance are those subgraphs that are also representable as a hierarchical product. Definition 4. Let G = G 1 (U 1 ) G n 1 (U n 1 ) G n, where U i V (G i ) for 1 i n 1. A subproduct in G is a subgraph of the form H 1 (W 1 ) H n 1 (W n 1 ) H n, where H i G i for 1 i n and W j U j for 1 j n 1. Given a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n and any two incident edges e and f in E(G), we say that e and f are in different layers if for some 1 j < k n we have p j (e) E(G j ) and p k (f) E(G k ). Given any two incident edges of G that are contained in different layers, these edges belong to exactly one of two unique structures in G. Definition 5. Let G = G 1 (U 1 ) G n 1 (U n 1 ) G n, where U i V (G i ) for 1 i n 1. Given some vertex x i V (G i ) for each 1 i n and vertices y j V (G j ) \ {x j } and y k V (G k ) \ {x k } for some 1 j < k n, consider the subgraph H of G induced by the vertices u 1 = (x 1,..., x j..., x k,..., x n ), u 2 = (x 1,..., y j,..., x k,..., x n ), u 3 = (x 1,..., y j,..., y k,..., x n ), u 4 = (x 1,..., x j,..., y k,..., x n ). We say that H is a semi-square if E(H) = {u 1 u 2, u 2 u 3, u 3 u 4 }. In this case, one can easily verify that y j U j, x j y j E(G j ), and x k y k E(G k ). We say that H is a square if E(H) = {u 1 u 2, u 2 u 3, u 3 u 4, u 1 u 4 }. In this case, it follows that x j, y j U j, x j y j E(G j ), and x k y k E(G k ). In Figure 4, we illustrate both a semi-square, in Figure 4(a), and a square, in Figure 4(b), for the graph P 3 (U 1 ) P 3. In both figures, the edges of the square or semi-square are in bold. Note that the following result is an immediate consequence of Definition 5.

6 (a) U 1 = {1} (b) U 1 = {0, 1} Figure 4. P 3 (U 1 ) P 3, where V (P 3 ) = {0, 1, 2} Lemma 6. Given a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n, where U i V (G i ) for 1 i n 1, let e and f be two incident edges that are in different layers; that is, e is in a G j -layer and f is in a G k -layer for some 1 j < k n. Then there exists exactly one semi-square or square in G containing e and f which has no diagonals. Proof. We may write e = uw = (x 1,..., x j,..., x k,..., x n )(x 1,..., y j,..., x k,..., x n ), where x i V (G i ) for 1 i n and y j V (G j ). Clearly, x j y j E(G j ) and x i U i for 1 i j 1. Similarly, we may write f = wv = (x 1,..., y j,..., x k,..., x n )(x 1,..., y j,..., y k,..., x n ), where y k V (G k ) and x k y k E(G k ). Moreover, y j U j and x i U i for j + 1 i k 1. Note that by Definition 5, any square or semi-square containing e and f must be unique. So we need only to show that e and f are indeed contained in a square or semi-square. We know g = zv = (x 1,..., x j,..., y k,..., x n )(x 1,..., y j,..., y k,..., x n ) E(G) since x i U i for 1 i j 1 and x j y j E(G j ). If x j U j, then uz E(G), and we may conclude that there exists a unique square H containing edges e, f, g, and uz. Otherwise, there exists a unique semi-square H containing edges e, f, and g. In either case, H does not contain a diagonal by definition of the edge set for this product. Given a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n, let H be a subgraph of G and G be the Cartesian product associated with G. Let a and b be two distinct vertices of V (H). Since a and b are vertices of V (G ) and G is connected, we let P represent a G -path from a to b. We now wish to determine when a subgraph of a hierarchical product can be written as a subproduct. Using the notion of unique squares and semi-squares, we define the following property that will identify such subgraphs.

7 Definition 7. Let H be a subgraph of G = G 1 (U 1 ) G n 1 (U n 1 ) G n, and let a and b be two distinct vertices of V (H). We say H is hierarchically convex if for any shortest G -path P between a and b, each G i -edge of P contained in G is also contained in H for 1 i n. For example, consider the hierarchical product G = P 3 (U 1 ) P 3, where V (P 3 ) = {0, 1, 2} and U 1 = {1, 2}. In Figure 5, we illustrate two subgraphs of G, H 1 and H 2, where the bold edges represent edges of H i for i {1, 2} and the dashed edges represent the missing edges of G contained in G. H 1, depicted in Figure 5(a), is not hierarchically convex, while H 2, depicted in Figure 5(b), is hierarchically convex. Consider the case when the hierarchical product G is in fact the Cartesian product; that is, G = G. Then to say a subgraph H is hierarchically convex reduces to the usual definition of convexity in that any shortest G-path between vertices a and b of V (H) is also contained in H. However, when G is not isomorphic to G, a subgraph H which is hierarchically convex may not be connected (a) H 1 is not hierarchically convex (b) H 2 is hierarchically convex Figure 5. P 3 (U 1 ) P 3 where V (P 3 ) = {0, 1, 2} and U 1 = {1, 2} Lemma 8. If H is a subgraph of a hierarchical product G = G 1 (U 1 ) G n 1 (U n 1 ) G n that is hierarchically convex, then for any two incident edges e and f of E(H) that are in different layers, the unique square or semi-square of G that contains e and f is also in H. Proof. Let G be the Cartesian product associated with G. Suppose there exist incident edges e and f of H that are in different layers. As in Lemma 6, we shall assume e is in a G j -layer and f is in a G k -layer for some 1 j < k n. Let x i U i for 1 i k 1 and x i V (G i ) for k i n. Write e = uw = (x 1,..., x j,..., x k,..., x n )(x 1,..., y j,..., x k,..., x n ), where x j y j E(G j ). Similarly, write f = wv = (x 1,..., y j,..., x k,..., x n )(x 1,..., y j,..., y k,..., x n ), where x k y k E(G k ). Letting z = (x 1,..., x j,..., y k,..., x n ), observe that P 1 = uwv and P 2 = uzv

8 8 are shortest G -paths from u to v. If u, w, v, and z are contained in a square in G, then the edges of this square are also contained in H since H is hierarchically convex. If u, w, v and z are contained in a semi-square in G, then x j U j and so uz E(H). However, zv is contained in G, so it is also contained in H. Thus, the unique square or semi-square in G containing e and f is also contained in H. Before we show that a subgraph H that is hierarchically convex in G is necessarily a subproduct, recall the distance between two vertices of a Cartesian product. Proposition 9 ([6], p. product G 1 G 2, then 51). If (x 1, x 2 ) and (y 1, y 2 ) are vertices of a Cartesian d G1 G 2 ((x 1, x 2 ), (y 1, y 2 )) = d G1 (x 1, y 1 ) + d G2 (x 2, y 2 ). Lemma 10. If G is a hierarchical product and H is a hierarchically convex subgraph of G, then H is a subproduct. Proof. Let H be a subgraph which is hierarchically convex in G. Let a and b be two distinct vertices of V (H). By associativity of the hierarchical product, it suffices to prove the lemma when G = G 1 (U 1 ) G 2. Write a = (a 1, a 2 ) and b = (b 1, b 2 ). We wish to show that (a 1, b 2 ) and (b 1, a 2 ) are also vertices of V (H). If a 1 = b 1, then any shortest G -path between a and b is of the form (a 1, a 2 )(a 1, x 1 2 ) (a 1, x k 2 )(a 1, b 2 ) where x i 2 V (G 2) for 1 i k. If a 1 U 1, then no edge of this path is contained in G and so the result is trivially true. If a 1 U 1, then every edge of this path is contained in G and consequently contained in H as well as H is hierarchically convex. So we may assume that a 1 b 1. If a 2 = b 2, then any shortest G -path between a and b is of the form (a 1, a 2 )(x 1 1, a 2) (x k 1, a 2)(b 1, a 2 ) where x i 1 V (G 1) for 1 i k. Each edge of this path is contained in G and therefore H. So we may assume that a 2 b 2. Let P be a shortest G -path between a and b. Every edge of P is mapped into a single vertex by one of the projections p 1 or p 2 and into an edge by the other. Proposition 9 implies that p 1 (P ) is a shortest path in G 1 from a 1 to b 1 and p 2 (P ) is a shortest path in G 2 from a 2 to b 2. Write p 1 (P ) = a 1 x 1 1 x2 1 xj 1 b 1, where x i 1 V (G 1) for 1 i j. Similarly, write p 2 (P ) = a 2 x 1 2 x2 2 xk 2 b 2, where x i 2 V (G 2) for 1 i k. Define {a 1 } p 2 (P ) = (a 1, a 2 )(a 1, x 1 2) (a 1, x k 2)(a 1, b 2 ) and p 1 (P ) {b 2 } = (a 1, b 2 )(x 1 1, b 2 ) (x j 1, b 2)(b 1, b 2 ). Note that the concatenation of {a 1 } p 2 (P ) and p 1 (P ) {b 2 } is a path of length d G1 (a 1, b 1 )+d G2 (a 2, b 2 ) making it a shortest G -path from (a 1, a 2 ) to (b 1, b 2 ). Furthermore, the path p 1 (P ) {b 2 } is contained in G, which implies this path is also contained in H since H is hierarchically convex. It follows that (a 1, b 2 ) V (H).

9 Similarly, we can define and {b 1 } p 2 (P ) = (b 1, a 2 )(b 1, x 1 2) (b 1, x k 2)(b 1, b 2 ) p 1 (P ) {a 2 } = (a 1, a 2 )(x 1 1, a 2 ) (x j 1, a 2)(b 1, a 2 ), where p 1 (P ) {a 2 } is contained in G. Thus, (b 1, a 2 ) V (H), and we may conclude that H is a subproduct. Notice that every layer G a j = {a 1 } G j {a n }, where a i U i for 1 i < j is a connected subproduct that is hierarchically convex in G. Theorem 11. Let φ be an isomorphism between the connected graphs G and H that are representable as hierarchical products G = G 1 (U 1 ) G n 1 (U n 1 ) G n and H = H 1 (W 1 ) H m 1 (W m 1 ) H m of prime graphs, where U i V (G i ) for i {1,..., n 1} and W j V (H j ) for j {1,..., m 1}. Then for some a V (G), there exists a permutation π of {1,..., n} such that φ(g a i ) = Hφ(a) π(i) for 1 i n and m = n. Proof. Fix j {1,..., n} and a = (a 1,..., a n ), where a i U i for 1 i < j. Assume that φ(a) = b = (b 1,..., b m ). As stated above, G a j is a connected subgraph of G that is hierarchically convex in G. Therefore, φ(g a j ) is connected and is hierarchically convex in H. By Lemma 10, φ(g a j ) is a subproduct; hence, we may write (b 1,..., b m ) φ(g a j ) = F 1 (U 1 ) F m 1 (W m 1 ) F m. Since G j = G a j = φ(g a j ) is prime, F i = {b i } for all except one i, 1 i j, which we denote as π(i). Moreover, φ(g a j ) is connected, so b i W i for each i ( ) {1,..., π(i) 1}. Therefore, φ(g a j ) F φ(a) π(i). This implies that Ga j φ 1 F φ(a) π(i). ( ) Since φ 1 F φ(a) π(i) is hierarchically convex in G, it is also a subproduct. Moreover, ( ) it is prime so φ 1 F φ(a) π(i) G a j. Thus, φ(ga j ) = F φ(a) π(i). We claim that the map π : {1, 2,..., n} {1, 2,..., m} is injective. If π(i) = π(j), then φ(g a i ) = F φ(a) π(i) = φ(ga j ). Since F φ(a) π(i) is nontrivial, it follows that G a i and G a j have a nontrivial intersection. This means i = j, so π is injective. Thus, n m. Repeating this argument for φ 1 gives m n, so n = m, and π is a permutation. 3. Preliminary Domination Bounds Arguably the most well-known conjecture in domination theory is Vizing s Conjecture from 1968 that states γ(gh) γ(g)γ(h) (see [3]). In light of the results in the previous section, we begin our study of the domination number in specific 9

10 10 hierarchical products. Our first observation is that given two graphs G and H, the domination number of the hierarchical product of G and H is monotonic with respect to nested root sets. Observation 12. Given any graph G of order n and any graph H, γ(g(u 1 ) H) γ(g(u 2 ) H) γ(g(u n ) H) = γ(gh), where U 1 U 2 U n and U k = k for 1 k n. Proof. This follows from the fact that any dominating set of G(U i ) H is also a dominating set of G(U i+1 ) H. Next, we determine bounds for the domination number when G is a path. We first consider when the root set contains only one vertex. Certainly if U =, then P m ( ) H is simply the disjoint union of V (H) paths, in which case γ(p m ( ) H) = V (H) γ(p m ). We now determine when γ(p m (U) H) = V (H) γ(p m ) when U contains a single vertex. We make use of the following notation given a hierarchical product G(U) H. Given v U, we let H v = {(v, y) y V (H)}. Given a set D V (G(U) H), we let π H (D) represent the projection of D onto the set V (H). That is, we let π H (D) = {v V (H) (x, v) D}. Theorem 13. Let G = P m and U = {x j } where 1 j m. For any graph H of order n, γ(p m (U) H) = nγ(p m ) if either m 1 (mod 3) or j 1 (mod 3). Otherwise, γ(p m (U) H) = n(γ(p m ) 1) + γ(h). Proof. We let (a 1, a 2 ) = (m (mod 3), j (mod 3)) represent the combination of congruence classes of m and j modulo 3. Since G( ) H is a subgraph of P m (U) H, we know that γ(p m (U) H) nγ(p m ) which shows the upper bound when (a 1, a 2 ) (1, 1). If (a 1, a 2 ) = (1, 1), then given any minimum dominating set D H of H, let D 1 = {(x 1, y) y D H }, and D 2 = {(x i, y) i 0 (mod 3), y V (H)}. One can easily verify that D 1 D 2 is a dominating set of P m (U) H of order γ(h) + n(γ(p m ) 1). Next, suppose that D is any dominating set of P m (U) H and let D x = D H xj. We first consider the projection of D x onto H. Let D = p H (D x ). Notice that for each y D, each vertex of V (G y ) {(x j 1, y), (x j, y), (x j+1, y)} is dominated from within its G-layer. It follows that D G y 1 + γ(p j 2 ) + γ(p m j 2 ). One can easily verify that 1 + γ(p j 2 ) + γ(p m j 2 ) γ(p m ) depending on (a 1, a 2 ).

11 If y D, then D G y γ(p j 1 ) + γ(p m j ). We leave it to the reader to verify that { γ(p m ) 1 if (a 1, a 2 ) = (1, 1) γ(p j 1 ) + γ(p m j ) = γ(p m ) otherwise. Therefore, if (a 1, a 2 ) (1, 1), we have D = D G y y V (H) D (1 + γ(p j 2 ) + γ(p m j 2 )) + (n D )(γ(p j 1 + γ(p m j )) D γ(p m ) + (n D )γ(p m ) = nγ(p m ). So assume that (a 1, a 2 ) = (1, 1). Let A = N H (D ) \ D and let B = V (H) \ (A D ). Note that if y B, then D G y γ(p m ). Moreover, D B is a dominating set of H so D + B γ(h). This implies that A n γ(h) and we have D = D G y y V (H) D γ(p m ) + B γ(p m ) + A (γ(p m ) 1) = nγ(p m ) A nγ(p m ) (n γ(h)) = n(γ(p m ) 1) + γ(h). We can use the above result to find a lower bound for the domination number of P m (U i ) H when we consider nested root sets as in Observation 12. In what follows, we let P 0 represent the empty graph. Theorem 14. For any m N, any s {1,..., m}, and any graph H of order n, where U s = {x 1,..., x s }. γ(p m (U s ) H) γ(h)γ(p s ) + (n γ(h))γ(p m s ), Proof. Let H be any fixed graph with order H. We proceed by induction on m with base cases m = 1, 2, or 3. Case 1 Suppose m = 1 and consider the graph P 1 (U 1 ) H. It follows that γ(p 1 (U 1 ) H) = γ(p 1 H) γ(h)γ(p 1 ). Case 2 Suppose m = 2. Notice that when s = 1, Theorem 13 guarantees that γ(p 2 (U 1 ) H) nγ(p 2 ) = γ(h)γ(p 1 ) + (n γ(h))γ(p 1 ). 11

12 12 On the other hand, when s = 2, we have γ(p 2 (U 2 ) H) = γ(p 2 H) γ(h) = γ(h)γ(p 2 ) + (n γ(h))γ(p 0 ). Thus, the result holds when m = 2. Case 3 Suppose m = 3. Notice that when s = 1, Theorem 13 guarantees that γ(p 3 (U 1 ) H) nγ(p 3 ) = (n γ(h))γ(p 3 ) + γ(h)γ(p 3 ) = γ(h)γ(p 1 ) + (n γ(h))γ(p 2 ). Next, assume that s = 2 and consider the graph P 3 (U 2 ) H. Let D be any dominating set of P 3 (U 2 ) H and set D = p H (D H x2 ). Note that if y D, then D G y 1 and if y D, then (x 3, y) D. Therefore, in both cases we know D G y 1 so that D = D G y y V (H) n = (n γ(h)) + γ(h) = (n γ(h))γ(p 1 ) + γ(h)γ(p 2 ). Finally, assume that s = 3 and consider the graph P 3 (U 3 ) H. Then γ(p 3 (U 3 ) H) = γ(p 3 H) γ(h) = γ(h) 1+(n γ(h)) 0 = γ(h)γ(p 3 )+ (n γ(h))γ(p 0 ). Suppose for some fixed k and all m {1,..., k} that the statement of the theorem is true. Consider the graph P m (U s ) H where m = k + 1. We proceed by induction on s with the base case s = 1. From Theorem 13, we know that if m 0, 2 (mod 3), then γ(p m (U 1 ) H) nγ(p m ) Similarly, if m 1 (mod 3), then = (n γ(h))γ(p m ) + γ(h)γ(p m ) = (n γ(h))γ(p m 1 ) + γ(h)γ(p m ) (n γ(h))γ(p m 1 ) + γ(h)γ(p 1 ). γ(p m (U 1 ) H) n(γ(p m ) 1) + γ(h) = nγ(p m 1 ) + γ(h)γ(p 1 ) (n γ(h))γ(p m 1 ) + γ(h)γ(p 1 ). So assume that for some fixed j {1,..., m 1} that for all 1 s j, we have γ(p m (U s ) H) γ(h)γ(p s ) + (n γ(h))γ(p m s ). Consider when s = j + 1 and let D be any dominating set of P m (U s ) H. Let D = p H (D H xs ), let A = N H (D ) \ D, and let B = V (H) \ (A D ).

13 First, assume that s = m. Thus, m > 3 and we may define G to be the subgraph of P m (U s ) H induced by the set { (x i, y) 1 i s 3, y V (H) }. If y D, then D G y D G y + D {(x s 2, y), (x s 1, y)} + 1. If y A, then D G y D G y + D {(x s 2, y), (x s 1, y)}. If y B, then D G y D G y + D {(x s 2, y)} + 1. Thus, D = D G y y V (H) D G + D H xs 2 + D H xs 1 + D γ(p s 3 H) + B + D. Since D B is a dominating set of H, B + D γ(h) and D γ(p s 3 H) + γ(h). Finally, note that s 3 = j 2 and by the inductive hypothesis, we know that for m = j 2 and U j 2, γ(p j 2 (U j 2 ) H) = γ(p j 2 H) γ(h)γ(p s 3 )+(n γ(h))γ(p 0 ) = γ(h)γ(p s 3 ). 13 Therefore, D γ(p s 3 H) + n γ(h)γ(p s 3 ) + n γ(h)(γ(p m ) 1) + 2γ(H) γ(h)γ(p m ). Next, assume that s < m and let G be the graph induced by the set Notice that if y D, then { (x i, y) 1 i s 2, y V (H) }. D G y D G y + D {(x s 1, y)} + γ(p m s 1 ) + 1 D G y + D {(x s 1, y)} + γ(p m s ). On the other hand, if y D, then D G y D G y + D {(x s 1, y)} + γ(p m s ). Summing over all y V (H), we have D D G y + D {(x s 1, y)} + γ(p m s ) y V (H) = D G + D H xs 1 + nγ(p m s ) γ(p s 2 H) + nγ(p m s ).

14 14 Notice that s 2 = j 1 and by the inductive hypothesis, we know that for m = j 1 and U j 1 that γ(p j 1 (U j 1 ) H) = γ(p j 1 H) γ(h)γ(p s 2 )+(n γ(h))γ(p 0 ) = γ(h)γ(p s 2 ). Therefore, D γ(h)γ(p s 2 ) + nγ(p m s ) = γ(h)γ(p s 2 ) + (n γ(h))γ(p m s ) + γ(h)γ(p m s ) γ(h) (γ(p s ) 1) + (n γ(h))γ(p m s ) + γ(h) = γ(h)γ(p s ) + (n γ(h))γ(p m s ). We would like to point out that although Jacobson and Kinch [7], and independently, Meir and Moon [10] proved γ(p m H) γ(p m )γ(h) for any m N, the above argument includes a unique proof for this same result. This further emphasizes the utility in studying domination within the hierarchical product as a means to understanding domination in the Cartesian product. We also note that similar arguments may be used when considering P m (U) H where U induces a disconnected graph. Finally, we would like to give a general lower bound for γ(g(u) H) when G and H are any two graphs. The following lemma allows us to partition V (G) prior to considering the hierarchical product. Lemma 15. Let G be any graph with γ(g) = k. There exists a partition A 1,..., A k of V (G) such that (i) A i is a clique for 1 i k 1, (ii) for each v A k, there exists w A i such that vw E(G) for 1 i k 1, and (iii) for each v A i 1 i k 1, there exists w A k such that vw E(G). Proof. Let D = {x 1,..., x k } be a γ-set of G. Choose a maximal clique from N[x 1 ] \ (D {x 1 }) that contains x 1 and call it A 1. Next, choose a maximal clique from N[x 2 ] \ (D {x 2 } A 1 ) that contains x 2 and call it A 2. Continue this process until A 1,..., A k 1 have been chosen and let A k = V (G) \ ( k 1 i=1 A i). Note that for some 2 i k 1, A i may only contain the vertex x i. Let w A i for some 1 i k 1. If w is adjacent to every vertex of A k, then j {i,k} {x j} {w} is a dominating set of G, contradicting γ(g) = k. So there exists some v A k such that wv E(G). On the other hand, if there exists u A k that is adjacent to every vertex of A i for some 1 i k 1, then A i was not a maximal clique. Thus, there exists some v A i such that uv E(G). Theorem 16. Let G be any graph with γ(g) = 3, and let A 1, A 2, A 3 be any partition of V (G) satisfying the conditions of Lemma 15. For any graph H of order n, γ(g(u) H) 2n where U A 1 or U A 2.

15 Proof. Assume that U A 1. Let D be any minimum dominating set of G(U) H. Let E represent the vertices of π H (D) that have exactly two pre-images in D. First note that if v V (H) \ π H (D), then each vertex of {(x, v) x A 3 } is not dominated by D, which cannot happen. So π H (D) = V (H). Suppose there exists v π H (D) where v has only one pre-image in D, call it (x, v). If x A 1, then there exists some vertex w A 2 A 3 that is not adjacent to x for otherwise γ(g) = 1. However, this implies D does not dominate (w, v) which is a contradiction. If x A 3, then by the choice of A 1, A 2, and A 3, there exists w A 2 that is not adjacent to x, another contradiction. If x A 2 and x is adjacent to each vertex of A 3, then {x, z} dominates G for any z A 1. However, this contradicts γ(g) = 3 so there exists w A 3 that is not adjacent to x. It follows that for each y V (H), D G y 2. Therefore, D 2 E + 3 V (H) \ E 2n. A similar argument shows the same result holds when U A 2. Theorem 17. Let G be any graph with γ(g) = 4, and let A 1, A 2, A 3, A 4 be any partition of V (G) satisfying the conditions of Lemma 15. For any graph H of order n, γ(g(u) H) 3n where U A i for some 1 i Proof. Assume U A 1 and let D be any minimum dominating set of G(U) H. Let E represent the vertices of π H (D) that have exactly two pre-images in D. As in the proof of Theorem 16, π H (D) = V (H) and no vertex of π H (D) has exactly one preimage in D. Suppose that v E and let (x, v) and (y, v) represent the pre-images of v in D. Note that {x, y} must dominate G \ A 1 since each vertex of the form (w, v) where w A 2 A 3 A 4 is dominated by either (x, v) or (y, v). However, this implies that {a, x, y} dominates G for any a A 1 as A 1 is a clique. This contradiction shows that no such v E exists so E =. It follows that D 3n. A similar argument shows the same result holds when U A 2 or U A 3. We point out that the above argument can be generalized for any graph G with domination number k 5. Yet there exist many open questions regarding domination in the hierarchical product. For instance, given two graph G and H where γ(g) = 3, can we partition V (G) into three sets A 1, A 2, A 3 in a slightly different way in order to obtain a lower bound for G(U) H when U is not restricted to precisely one of A i? In general, can we find a lower bound for G(U) H regardless of the structure of G or choice of U? Is it possible to determine the gap between γ(gh) and γ(g(u) H) based on the choice of U? References [1] M. Azari and A. Iranmanesh. Chemical graphs constructed from rooted products and their Zagreb indices. MATCH Commun. Math. Comput. Chem. 70: , [2] L. Barrière, C. Dalfó, M.A. Fiol, and M. Mitjana. The generalized hierarchical product of graphs. Discrete Math., 309(12): , 2009.

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